A thin spherical shell of metal has a radius of 0.25 m and carries a chaarge of 0.2 mC. The electric indtensity at a point for outside the shell will be

Options

A. `2.88 xx 10^(4) N//C`
B. `3.4 xx 10^(4) N//C`
C. `325 xx 10^(4) N//C`
D. `3.88 xx 10^(4) N//C`

Answer

To find the electric field intensity at a point outside a thin spherical shell of metal, we can use Gauss's law. According to Gauss's law, the electric field (E) outside a uniformly charged spherical shell behaves as if all the charge were concentrated at the center of the shell.

### Step-by-Step Solution:

1. **Identify the given values**:
- Radius of the spherical shell (R) = 0.25 m
- Charge on the shell (Q) = 0.2 mC = 0.2 × 10^(-6) C

2. **Use the formula for electric field due to a point charge**:
The electric field (E) at a distance (r) from a point charge (Q) is given by the formula:
\[ E = \frac{k \cdot Q}{r^2} \]
where \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Choose a point outside the shell**:
Since we are interested in the electric field at a point outside the shell, we can choose a distance \( r \) greater than the radius of the shell. For example, let’s take \( r = 0.5 \, \text{m} \) (which is greater than 0.25 m).

4. **Substitute the values into the formula**:
\[ E = \frac{9 \times 10^9 \cdot (0.2 \times 10^{-6})}{(0.5)^2} \]

5. **Calculate \( (0.5)^2 \)**:
\[ (0.5)^2 = 0.25 \]

6. **Now substitute back into the equation**:
\[ E = \frac{9 \times 10^9 \cdot (0.2 \times 10^{-6})}{0.25} \]

7. **Calculate \( 9 \times 10^9 \cdot 0.2 \times 10^{-6} \)**:
\[ 9 \times 10^9 \cdot 0.2 \times 10^{-6} = 1.8 \times 10^4 \]

8. **Now divide by 0.25**:
\[ E = \frac{1.8 \times 10^4}{0.25} = 7.2 \times 10^4 \, \text{N/C} \]

9. **Final result**:
The electric field intensity at a point outside the shell is:
\[ E = 7.2 \times 10^4 \, \text{N/C} \]